CBSE Class 10 Answered
(sin70/cos20)+(cos36/sec54)-(2cos43cos47/tan10tan40tan80tan10)
Asked by sanskriti | 19 Sep, 2013, 12:27: PM
Expert Answer
Pls cross check your question, it should be as written below.
(sin70/cos20)+(cosec36/sec54)-(2cos43cosec47/tan10tan40tan50tan10)
= sin[90-20] / [cos20] + [cosec36] / sec[90-36] - {2cos43 cosec [90-43] / tan10. tan40. tan[90-40]. tan[90-10]}
=[cos20] / [cos20] + [cosec36 / cosec36] - [2cos43sec43 / tan10. tan40. cot40. cot10]
= 1 + 1 - 2
= 0
Answered by | 19 Sep, 2013, 12:36: PM
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